Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The matrix is $ M= \frac{d}{d\theta} e^{A+\theta B} \mid _{\theta = 0} $ where $A$ and $B$ are both $n\times n$ matrices. I was thinking solving it by introducing the equations: $\dot x = (A + \theta B)x,\ x(0) = I$ with solution $x = X(t,\theta) $, where $M = \frac{dX(1,0)}{d\theta}$, I was stuck then, thanks for any suggestions : )

share|cite|improve this question
Welcome to Math.SE! Try the Taylor series for $\exp$: the term of 1st degree in $\theta$ should be exactly what you need. – user53153 Jan 23 '13 at 1:37
up vote 1 down vote accepted

To flesh out 5PMs comment: $$\exp(A + \theta B) = \sum_{k \geqslant 0} \frac{(A + \theta B)^k}{k!} = \sum_{k \geqslant 0} \frac{A^k + k A^{k-1}(\theta B) + \theta^2( \ldots )}{k!} $$Differentiating term by term, we get $$\sum_{k \geqslant 0} \frac{k A^{k-1} B + \theta( \ldots)}{k!}$$So when we evaluate at zero, we obtain: $$\sum_{k \geqslant 0} \frac{A^{k-1}}{(k-1)!} B = \exp(A)B$$

share|cite|improve this answer
Hmm, this works only when $A$ and $B$ commute. – user1551 Jan 23 '13 at 13:24
hey man, yeah, I goofed that up. However, it nearly works, instead you get $\sum_{k \geqslant 0} \frac{1}{k!} (\sum_{j=0}^{k-1} A^j B A^{k-j})$ – uncookedfalcon Jan 23 '13 at 20:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.